Term Split By Hashtag Of Multiple Words

I am trying to split a term which contains a hashtag of multiple words such as '#I-am-great' or '#awesome-dayofmylife' then the output that I am looking for is: I am great aweso

Solution 1:

All the commentators above are correct of course: A hashtag without spaces or other clear separators between the words (especially in English) is often ambiguous and cannot be parsed correctly in all cases.

However, the idea of the word list is rather simple to implement and might yield useful (albeit sometimes wrong) results nevertheless, so I implemented a quick version of that:

wordList = '''awesome day of my life because i am great something some
thing things unclear sun clear'''.split()

wordOr = '|'.join(wordList)

defsplitHashTag(hashTag):
  for wordSequence in re.findall('(?:' + wordOr + ')+', hashTag):
    print':', wordSequence   
    for word in re.findall(wordOr, wordSequence):
      print word,
    printfor hashTag in'''awesome-dayofmylife iamgreat something
somethingsunclear'''.split():
  print'###', hashTag
  splitHashTag(hashTag)

This prints:

### awesome-dayofmylife
: awesome
awesome
: dayofmylife
day of my life
### iamgreat
: iamgreat
i am great
### something
: something
something
### somethingsunclear
: somethingsunclear
something sun clear

And as you see it falls into the trap qstebom has set for it ;-)

EDIT:

Some explanations of the code above:

The variable wordOr contains a string of all words, separated by a pipe symbol (|). In regular expressions that means "one of these words".

The first findall gets a pattern which means "a sequence of one or more of these words", so it matches things like "dayofmylife". The findall finds all these sequences, so I iterate over them (for wordSequence in …). For each word sequence then I search each single word (also using findall) in the sequence and print that word.

Solution 2:

The problem can be broken down to several steps:

1. Populate a list with English words
2. Split the sentence into terms delimited by white-space.
3. Treat terms starting with '#' as hashtags
4. For each hashtag, find words by longest match by checking if they exist in the list of words.

Here is one solution using this approach:

# Returns a list of common english terms (words)definitialize_words():
    content = Nonewithopen('C:\wordlist.txt') as f: # A file containing common english words
        content = f.readlines()
    return [word.rstrip('\n') for word in content]


defparse_sentence(sentence, wordlist):
    new_sentence = ""# output    
    terms = sentence.split(' ')    
    for term in terms:
        if term[0] == '#': # this is a hashtag, parse it
            new_sentence += parse_tag(term, wordlist)
        else: # Just append the word
            new_sentence += term
        new_sentence += " "return new_sentence 


defparse_tag(term, wordlist):
    words = []
    # Remove hashtag, split by dash
    tags = term[1:].split('-')
    for tag in tags:
        word = find_word(tag, wordlist)    
        while word != Noneandlen(tag) > 0:
            words.append(word)            
            iflen(tag) == len(word): # Special case for when eating rest of wordbreak
            tag = tag[len(word):]
            word = find_word(tag, wordlist)
    return" ".join(words)


deffind_word(token, wordlist):
    i = len(token) + 1while i > 1:
        i -= 1if token[:i] in wordlist:
            return token[:i]
    returnNone 


wordlist = initialize_words()
sentence = "big #awesome-dayofmylife because #iamgreat"
parse_sentence(sentence, wordlist)

It prints:

'big awe some day of my life because i am great '

You will have to remove the trailing space, but that's easy. :)

I got the wordlist from http://www-personal.umich.edu/~jlawler/wordlist.